Question

A fighter plane is flying horizontally at an altitude of $1.5km$ with speed $720km{h}^{-1}$. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target? (Take $g=10m{s}^{-2}$)

• A. $ta{n}^{-1}\left(\frac{\sqrt{3}}{4}\right)$
• B. $ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• C. $ta{n}^{-1}\left(\frac{\sqrt{1}}{2}\right)$
• D. $ta{n}^{-1}(2)$

Answer 1

Answer: (A)

Here, $u=720km{h}^{-1}$
$=720\times \frac{5}{18}m{s}^{-1}=200m{s}^{-1}$
$H=1.5km=1.5\times 1000m=1500m$
Time taken by the bomb to attack the target
$t=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\times 1500m}{10m{s}^{-1}}}$
$=\sqrt{300}s=10\sqrt{3}s$
$R=u\times t=200m{s}^{-1}\times 10\sqrt{3}s$
$=2000\sqrt{3}m$
From figure,
Angle of sight (w. r. t horizontal) $\alpha ={\tan }^{-1}\left(\frac{H}{R}\right)$
$\alpha ={\tan }^{-1}\left(\frac{1500}{2000\sqrt{3}}\right)={\tan }^{-1}\left(\frac{3}{4\sqrt{3}}\right)$
$={\tan }^{-1}\left(\frac{\sqrt{3}}{4}\right)$